Transcript Kinetics Notes- Part 2 - Kentucky Department of Education

Kinetics Notes- Part 2
TWO TYPES OF RATE LAWS
• Thus far, we have examined
Differential rate laws—
– data table contains concentration and rate data.
– Use table logic or algebra to determine the orders
of reactants and the value of the rate constant, k.
– Rate=k[A]m[B]n
• Differential rate laws can be converted (by use
of yucky calculus) into equations that show the
relationship between concentrations and time
known as
Integrated rate laws—
– data table contains concentration and time data.
– Use graphical methods to determine the order of a given
reactant.
• integrated rate laws for reactions that contain a single
reactant are as follows:
– zero order
– first order
– second order
y=
m x + b
↓
↓↓ ↓
[A] = −k t + [Ao]
↓
↓↓ ↓
ln[A]= −k t + ln [Ao]
↓
↓↓ ↓
1/[A] = k t + 1/[Ao]
• t is time; [A] is the concentration of reactant A at time t;
[A]0 is the concentration of A at time=0; and k is the rate
constant
• Notice that each of the integrated rate laws can be
arranged into the equation of a straight line, y=mx+b
– The value of the rate constant k is equal to
the absolute value of the slope of the best fit
line
• Requires you to make 3 graphs ([A] vs time),
(ln[A] vs time), and (1/[A] vs time) –to decide if
the reaction is 0th, 1st, or 2nd order
• the graph that gives the straight line is decided
by performing 3 linear regressions (on the zero,
first, and second order graphs) and analyzing the
regression correlation coefficient r. Not nearly as
hard as it sounds!
INTEGRATED RATE LAW: CONCENTRATION/TIME
RELATIONSHIPS
• When we wish to know how long a reaction
must proceed (time, t) to reach a specific
concentration of some reagent ([A]) , we can
determine the integrated rate law, and use it
because it gives a relationship between
concentration and time.
When given concentration and time
data, make your graphs!
•
Set up your axes so that time is
always on the x-axis.
•
Plot the concentration of the reactant
on the y-axis of the first graph.
•
Plot the natural log of the
concentration (ln [A], NOT log[A]) on
the y-axis of the second graph
•
and the reciprocal of the
concentration on the y-axis of the
third graph.
•
You are in search of linear data!
Which one gives you a straight line?
(you are looking for the graph that
gives you r=l1l), or the closest one to
it)
•
Here comes the elegant part… If you
do the set of graphs in this order with
the y-axes being
– “concentration”,
– “natural log of concentration” and
– “reciprocal concentration”,
the alphabetical order of the y-axis
variables leads to
– 0th ,
– 1st and
– 2nd
orders respectively for that reactant.
Zero order
k = negative slope
First order
k = negative slope
Second order
k = the slope
• You can now easily solve for either time or
concentration once you know the order of the
reactant.
• Just remember y = mx + b.
• Choose the set of variables that gave you the best
straight line (r value closest to ±1) and insert
them in place of x and y in the generalized
equation for a straight line.
• “A” is reactant A and Ao is the initial concentration
of reactant A at time zero [the y-intercept].
y
=
• zero order
[A] =
• first order
ln[A] =
• second order 1/[A] =
mx +
−kt +
−kt +
kt +
b
[Ao]
ln [Ao]
1/[Ao]
• Also recognize that lslopel= k, since the rate
constant is NEVER negative. If you are asked to
write the rate expression [or rate law] it is simply
Rate = k[A]order you determined from analyzing the graphs
Using the graphing calculator
• Set up your calculator so that time is always in L1.
Use L2, L3 and L4 to display the y-variables. Remember the list for what is
placed on the y-axis is alphabetical (concentration, natural log of
concentration and reciprocal concentration).
•
• L1 = time (x-variable for all three graphs)
• L2 = concentration
if [A] gives a straight line = zero order
• L3 = ln concentration
if ln [A] gives a straight line = first order
• L4 = reciprocal concentration if 1/[A] gives a straight line = second
order
• Use this system to set up the data given in the following exercise.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Exercise
The decomposition of N2O5 in the gas phase was studied at constant temperature.
2 N2O5(g) → 4 NO2(g) + O2(g)
The following results were collected:
[N2O5]
0.1000
0.0707
0.0500
0.0250
0.0125
0.00625
Determine the rate law and calculate the value of k.
What is the concentration of N2O5(g) at 600 s?
At what time is the concentration of N2O5(g) equal to 0.00150 M ?
Time (s)
0
50
100
200
300
400
• We are going to perform 3 linear regressions to
determine the order of the reactant. They will be
L1,L2; L1,L3; L1,L4.
• Next, we will determine which regression has the
best r-value [linear regression correlation
coefficient in big people language!]
• We will also paste the best regression equation
“Y=“ so that we can easily do other calculations
commonly required on AP Chemistry Exam
problems.
•
HALF-LIFE AND REACTION RATE FOR
FIRST ORDER REACTIONS, t1/2
• the time required for one half of one of the
reactants to disappear.
• t1/2 = 0.693
k
• Exercise
• A certain first-order reaction has a half-life of 20.0
minutes.
• a. Calculate the rate constant for this reaction.
•
•
•
• b. How much time is required for this reaction to be
75% complete?
•
•
• Exercise
• The rate constant for the first order
transformation of cyclopropane to propene is
5.40 × 10−2/hr. What is the half-life of this
reaction? What fraction of the cyclopropane
remains after 51.2 hours? What fraction
remains after 18.0 hours?
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Exercise
For the reaction of (CH3)3CBr with OH-,
(CH3)3CBr + OH-  (CH3)3COH + BrThe following data were obtained in the laboratory.
TIME (s)
[(CH3)3CBr]
0
0.100
30
0.074
60
0.055
90
0.041
Plot these data as ln [(CH3)3CBr] versus time. Sketch your graph.
Is the reaction first order or second order? What is the value of the rate constant?
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Exercise 12.5
Butadiene reacts to form its dimer according to the equation
2 C4H6 (g)  C8H12 (g)
The following data were collected for this reaction at a given temperature:
[C4H6]
0.01000
0.00625
0.00476
0.00370
0.00313
0.00270
0.00241
0.00208
Time ( 1 s)
0
1000
1800
2800
3600
4400
5200
6200
What is the order of this reaction? Explain. Sketch your graph as part of your explanation. Write the rate law expression:
What is the value of the rate constant for this reaction?
c. What if the half-life for the reaction under the conditions of this experiment?
HALF-LIFE AND REACTION RATE FOR
ZERO and SECOND ORDER REACTIONS
• t1/2 =
[A]o
2k
for a ZERO order rxn.
t1/2 = ____1____ for a SECOND order rxn.
k[A]0
• Zero-order reactions are most often
encountered when a substance such as a
metal surface or an enzyme is required for the
reaction to occur. The enzyme or catalyst may
be come saturated and therefore an increase
in the [reactant/substrate] has no effect on
the rate.
INTEGRATED RATE LAWS FOR REACTIONS WITH MORE
THAN ONE REACTANT
• Must still be determined by experiment! But we use a
technique called “swamping”.
• Flood the reaction vessel with high concentrations of
all but one reactant and perform the experiment. The
reactants at high concentrations like say, 1.0 M
compared to the reactant with a low concentration say,
1.0 x 10-3 M, stay the same.
• “In English”—the rate is now dependent on the
concentration of the little guy since the big guy’s aren’t
changing (much), therefore the rate = k’ [little guy]
• We now re-write the rate as a pseudo-rate-law and k’
is a pseudo-rate-constant
A summary
Graphs and kinetic data
1. Plotting the concentration of a reactant against time.
(i) a straight-line graph shows
zero order w.r.t that reactant.
Remember that one way to
define the rate is to use the
expression :
Rate= decrease in [reactant]/time
and so the slope of such a
graph will equal the rate.
Since the slope of the graph is
constant , i.e. the rate is
constant, we can say that
changing the concentration of
the reactant is having no effect
on the rate, i.e. it is zero order
w.r.t that reactant.
(ii) a constant half-life graph
shows first order w.r.t
that reactant.
• The graph shows constant
half lives and therefore is
first order w.r.t H2O2. The
rate constant can be
calculated using the
expression:
k= 0.693
t1/2
2. Plotting initial rates against
concentrations of the reactants.
Summary of graphical interpretations
and orders of reaction
[A]0
2k
1 .
k[A]0